The area of the conference table in mr. nathan’s office must be no more than 175 ft2. if the length of the table is 18 ft more than the width, x, which interval can be the possible widths?

The answer is  0 < x ≤ 7 

First, we know that width =  x
Which means that length = x +18

So, the possible equation for the Table's area is

X (X + 18)  ≤ 175

X^2 + 18x - 175  ≤ 0

Next, we need to calculate is by using complete square method
x^2 + 18x + 81 ≤ 175 + 81

(x + 9)^2 ≤ 256

|x + 9| ≤ sqrt(256)

|x + 9| ≤ +-16

-16 ≤ x + 9 ≤ 16

-16 - 9 ≤ x ≤ 16 - 9

-25 ≤ x ≤ 7

Since the width couldn't be negative, we can change -25 with 0,

so it become
 0 < x ≤ 7